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  • [Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.2.10

    (1). The numerical radius defines a norm on $scrL(scrH)$.

     

    (2). $w(UAU^*)=w(A)$ for all $Uin U(n)$.

     

    (3). $w(A)leq sen{A}leq 2w(A)$ for all $A$.

     

    (4). $w(A)=sen{A}$ if (but not only if) $A$ is normal.

     

    Solution.

     

    (1). We only need to show that $$eex ea w(A)=0& a sef{x,Ax}=0,quad forall x:sen{x}=1\ & a sef{y,Ax}=frac{1}{4} sum_{k=0}^3 i^ksef{x+i^ky,A(x+i^ky)}=0,quadforall x,y:sen{x}=sen{y}=1\ & a Ax=0,quad forall x:sen{x}=1\ & a A=0. eea eeex$$

     

    (2). $$eex ea w(UAU^*)&=sup_{sen{x}=1}sev{sef{x,UAU^*}}\ &=sup_{sen{x}=1}sev{(U^*x)^*A(U^*x)}\ &=sup_{sen{y}=1}sev{y^*Ay}quadsex{y=U^*x}\ &=w(A). eea eeex$$

     

    (3). $$eex ea w(A)&=sup_{sen{x}=1}sev{sef{x,Ax}}\ &leq sup_{sen{x}=1} sex{sen{x}cdot sen{Ax}}\ &=sup_{sen{x}=1}sen{Ax}\ &=sen{A};\ sen{A}&=sup_{sen{x}=sen{y}=1}sev{sef{y,Ax}}\ &=sup_{sen{x}=sen{y}=1} sev{frac{1}{4}sum_{k=0}^3 i^ksef{y+i^kx,A(y+i^kx)}}\ &leq sup_{sen{x}=sen{y}=1} frac{1}{4}sum_{k=0}^3 sev{sef{y+i^kx,A(y+i^kx)}}\ &leq sup_{sen{x}=sen{y}=1} frac{1}{4}sum_{k=0}^3 sen{y+i^kx}^2cdot w(A)\ &=sup_{sen{x}=sen{y}=1} frac{1}{4}cdot 4sex{sen{x}^2+sen{y}^2} cdot w(A)\ &=2w(A). eea eeex$$

     

    (4). If $A$ is normal, then by the spectral theorem, there exists a unitary $U$ such that $$ex A=Udiag(lm_1,cdots,lm_n)U^*, eex$$ and hence $$eex ea sen{Ax}^2&=sef{Ax,Ax}\ &=x^*A^*Ax\ &=Ux^*diag(|lm_1|^2,cdots,|lm_n|^2)U^*x\ &=sum_{i=1}^n |lm_i|^2|y_i|^2quadsex{y=U^*x}\ &leq max_isen{lm_i}^2sen{y}^2\ &leq w(A)sen{x}^2. eea eeex$$

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  • 原文地址:https://www.cnblogs.com/zhangzujin/p/4106633.html
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